hw2_soln - Physics 313 Problem Set 2 Due Friday September...

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Unformatted text preview: Physics 313 Problem Set 2 Due Friday September 22, 2006 by the end of class 1. Schroeder Problem 1.36 A. Use V P =constant to get (1 L ) (1 atm ) = V (7 atm ), so V = (1 / 7) (1 / ) . Here = 1 . 4 for air, so V = 0 . 249 L. B. To get the work, do W = R P ( V ) dV . Now V P =constant = 1 to start with (in units of literxs and atmospheres), so P = V- . W =- Z V f V i dV V- = 1 - 1 V- +1 f- V- +1 i = 1 . 4 ( . 249- . 4- 1- . 4 ) = 1 . 86 So W = 1 . 86 L atm, which is 188 J. C. I assume the initial temperature is 300 K. The volume decreased from 1 L to 0.249 L, but the pressure increased by a factor of 7. Using the ideal gas law, the final temperature should scale with P and V, and so T = (300 K )(0 . 249)(7) = 523 K. 1 2. Schroeder Problem 1.48 The incident power is 1000 W/m 2 , but 90% is reflected, so the absorbed power is 100 W/m 2 . The snow pack is 2 m thick. Imagine a 1 m 2 patch of the snow. Were told its 50% air, so the total volume of ice to be melted is (2)(0 . 5) = 1 m 2 = 10 6 cm 2 . Ice has a densityof 0.9, so this is 0 . 9 10 6 = 9 10 5 grams of ice to melt. Ice has a latent heat of 333 J/g, so melting all of the ice would require (333)(9 10 5 ) = 3 10 8 J. The ice pack would therefore last for (3...
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This note was uploaded on 03/11/2010 for the course PHYSICS 108 taught by Professor 0000 during the Fall '07 term at Clarke University.

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hw2_soln - Physics 313 Problem Set 2 Due Friday September...

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